Deriving the Ratio-Correction Formula

Why a Ratio Correction Appears

A DBCD uses information about two different quantities: the target allocation and the allocation already achieved. The target gives the desired share for an arm; the current share records how far the trial’s assignments have moved toward that target. The ratio-correction formula combines them to determine the probability of assigning the next participant.

This tutorial derives the two-arm formula from its weighting construction. The derivation is algebraic, but the resulting probability is not simply the target share: it depends on the target-to-current ratios in both arms. The distinction matters when an arm is underrepresented or overrepresented relative to its target.

Definition: Immediately before participant $n+1$ is assigned, let $N_{n,1}$ and $N_{n,2}$ be the numbers assigned to arms 1 and 2, with $N_{n,1}+N_{n,2}=n$. The current allocation shares are $q_{n,1}=N_{n,1}/n$ and $q_{n,2}=N_{n,2}/n=1-q_{n,1}$. Let $\rho_n$ be the current target share for arm 1, so the target share for arm 2 is $1-\rho_n$. The ratio-correction DBCD probability for arm 1 is $$ P(A_{n+1}=1\mid\mathcal F_n) = \frac{\rho_n\left(\rho_n/q_{n,1}\right)^\gamma} {\rho_n\left(\rho_n/q_{n,1}\right)^\gamma+ (1-\rho_n)\left((1-\rho_n)/q_{n,2}\right)^\gamma}, $$ where $\mathcal F_n$ represents the information available before the next assignment and $\gamma\geq 0$ is the correction exponent. The probability for arm 2 is one minus this probability.

The formula is defined directly when $n>0$, $0<\rho_n<1$, $0<q_{n,1}<1$, and $0<q_{n,2}<1$. These conditions ensure that each target and current share in a ratio is positive. In a sequential trial, an initial allocation procedure can ensure both arms have at least one assignment before the formula is used. The target may be based on response information, but the algebra below does not require a particular method for estimating it.

Deriving the Two-Arm Formula

Begin by assigning each arm an unnormalized weight. Its weight is its target share multiplied by its target-to-current share ratio raised to $\gamma$. For arm 1 this weight is $w_{n,1}=\rho_n(\rho_n/q_{n,1})^\gamma$; for arm 2 it is $w_{n,2}=(1-\rho_n)((1-\rho_n)/q_{n,2})^\gamma$. The allocation probability is obtained by dividing each weight by their sum.

Write the target and current shares.
Arm 1 has target share $\rho_n$ and current share $q_{n,1}$; arm 2 has target share $1-\rho_n$ and current share $q_{n,2}$.

Form the ratio corrections.
For each arm, divide its target share by its current share, then raise that ratio to $\gamma$. Multiply the result by the arm’s target share to obtain its weight.

Normalize the weights.
Divide arm 1’s weight by the sum of both weights. This makes the two assignment probabilities nonnegative and ensures they sum to one.

w_{n,1}=\rho_n\left(\frac{\rho_n}{q_{n,1}}\right)^\gamma,\qquad w_{n,2}=(1-\rho_n)\left(\frac{1-\rho_n}{q_{n,2}}\right)^\gamma

Thus the probability of assigning the next participant to arm 1 is $w_{n,1}/(w_{n,1}+w_{n,2})$, which gives the definition above. Since both weights are positive under the stated conditions, the probability lies strictly between zero and one. The probability for arm 2 is $w_{n,2}/(w_{n,1}+w_{n,2})$, and adding the two normalized weights gives one.

An Equivalent Odds Derivation

The same result can be expressed in terms of odds. Taking the ratio of the two assignment probabilities cancels the common normalizing denominator. The result separates the target odds from the relative ratio correction:

\frac{P(A_{n+1}=1\mid\mathcal F_n)} {P(A_{n+1}=2\mid\mathcal F_n)} = \frac{w_{n,1}}{w_{n,2}} = \frac{\rho_n}{1-\rho_n} \left[ \frac{\rho_n/q_{n,1}}{(1-\rho_n)/q_{n,2}} \right]^\gamma

This identity follows by substituting the two weights and collecting the target-share factors and correction factors separately. It says that the assignment odds equal the target odds multiplied by the relative correction factor raised to $\gamma$. The correction compares arm 1’s target-to-current ratio with arm 2’s target-to-current ratio; it is not just arm 1’s ratio in isolation.

Derivation: With positive weights, normalization gives $P(A_{n+1}=1\mid\mathcal F_n)=w_{n,1}/(w_{n,1}+w_{n,2})$ and $P(A_{n+1}=2\mid\mathcal F_n)=w_{n,2}/(w_{n,1}+w_{n,2})$. Dividing these expressions cancels the denominator, so the odds are $w_{n,1}/w_{n,2}$. Substituting the definitions of the weights yields the odds expression above. Conversely, for positive odds $R$, the corresponding arm 1 probability is $R/(1+R)$; substituting the displayed odds and simplifying recovers the normalized-weight formula. Therefore the weight and odds forms specify the same allocation probability.

Worked Calculations

Worked Example: Arm 1 Is Below Its Target Share

Suppose a hypothetical trial has $n=20$ participants: 8 assigned to arm 1 and 12 to arm 2. The current target share for arm 1 is $\rho_n=0.60$, and take $\gamma=1$. Find the probability of assigning participant 21 to arm 1.

The current shares are $q_{n,1}=8/20=0.40$ and $q_{n,2}=12/20=0.60$. Both current shares and both target shares are positive, so the formula is defined. The weights are

w_{n,1}=0.60\left(\frac{0.60}{0.40}\right)^1 =0.60(1.5)=0.900

For arm 2, the target share is $0.40$, giving

w_{n,2}=0.40\left(\frac{0.40}{0.60}\right)^1 =0.40(0.6667)=0.2667

Normalize the weights:

P(A_{21}=1\mid\mathcal F_{20}) =\frac{0.900}{0.900+0.2667} =0.7714

The next participant therefore has a 0.7714 assignment probability for arm 1 and a 0.2286 probability for arm 2. This probability is greater than arm 1’s 0.60 target share because its current share, 0.40, is below that target.

Worked Example: Arm 1 Is Above Its Target Share

In a second hypothetical trial, suppose $n=20$, with 14 assignments to arm 1 and 6 to arm 2. Let the target share for arm 1 be $\rho_n=0.40$, and let $\gamma=1$. Calculate the next assignment probability for arm 1.

Here $q_{n,1}=14/20=0.70$ and $q_{n,2}=6/20=0.30$. The counts in both arms are positive, and both target shares are strictly between zero and one. The weights are

w_{n,1}=0.40\left(\frac{0.40}{0.70}\right) =0.40(0.5714)=0.2286

$$ w_{n,2}=0.60\left(\frac{0.60}{0.30}\right) =0.60(2)=1.200 $$

Therefore,

P(A_{n+1}=1\mid\mathcal F_n) =\frac{0.2286}{0.2286+1.200} =0.1600

The probability of assigning the next participant to arm 1 is 0.1600, with probability $1-0.1600=0.8400$ for arm 2. Arm 1’s current share exceeds its 0.40 target, and its weight is correspondingly small relative to arm 2’s.

Worked Example: Unequal Corrections in the Same Allocation State

Consider hypothetical data with $n=12$, 3 participants assigned to arm 1, and 9 to arm 2. The target for arm 1 is $\rho_n=0.50$, and use $\gamma=2$. Compute the next assignment probability.

The current shares are $q_{n,1}=3/12=0.25$ and $q_{n,2}=9/12=0.75$. The required shares are positive, so the ratios can be calculated:

w_{n,1}=0.50\left(\frac{0.50}{0.25}\right)^2 =0.50(2^2)=2.000

$$ w_{n,2}=0.50\left(\frac{0.50}{0.75}\right)^2 =0.50\left(\frac{2}{3}\right)^2 =0.2222 $$

Normalizing gives

P(A_{n+1}=1\mid\mathcal F_n) =\frac{2.000}{2.000+0.2222} =0.9000

The next participant’s probability of assignment to arm 1 is 0.9000. This calculation illustrates how the exponent enters the weights: each target-to-current ratio is squared before normalization. The calculation does not require interpreting $\gamma$ as a probability or changing the target share.

Assumptions, Diagnostics, and Boundaries

The formula requires a target that is available at the time of assignment and current shares calculated from assignments made before that participant arrives. In a response-adaptive design, the target can depend on estimates based on observed responses; conditional on the information $\mathcal F_n$, however, the target used in the calculation must be specified. The next assignment is then randomized with the stated conditional probability.

Several checks help detect calculation or implementation errors:

Confirm that $q_{n,1}+q_{n,2}=1$ and that both shares use the same $n$.
Confirm that the arm 2 target is $1-\rho_n$, not another copy of $\rho_n$.
Check that each weight uses its own arm’s target share divided by its own current share.
After normalization, verify that both probabilities are between zero and one and sum to one.
Compare the result with the input allocation state: a very small arm-specific target-to-current ratio should not accidentally receive a large correction weight.

There are important boundary cases. If $n=0$, current shares defined as $N_{n,j}/n$ are undefined. If an arm has no assignments yet, its current share is zero and its ratio cannot be evaluated. If a target share is zero or one, some weights may vanish or the displayed ratios may be undefined. Such cases require an explicit initialization or boundary-handling rule; substituting zero into the interior formula is not a valid calculation. Any minimum allocation probabilities or other modifications must be specified as part of the design rather than silently assumed to be present here.

How the Formula Differs From Direct Target Allocation

Direct target allocation would set the next assignment probability for arm 1 equal to $\rho_n$, regardless of the observed allocation shares. Ratio correction instead starts with target-share weights and changes their relative sizes using the allocation state. If the current shares already match the targets, both arm-specific ratios equal one, so the weights reduce to the target shares and normalization gives $\rho_n$. Away from that condition, the allocation probability need not equal the target.

Approach	Arm 1 probability	Uses current allocation shares?
Direct target allocation	$\rho_n$	No
Ratio-correction DBCD	$w_{n,1}/(w_{n,1}+w_{n,2})$	Yes, through both arm-specific ratios

The ratio-correction formula defines a probability from the target and current shares; it does not by itself establish that the realized allocation will equal the target at any particular time. Random assignments can produce fluctuations, and the target itself may change when response information changes. Nor does the algebra establish that a particular target is ethically or statistically appropriate: target selection is a separate design decision.

Common Mistake

A frequent error is to calculate arm 1’s correction as $\rho_n/q_{n,1}$, multiply the target by that quantity, and then treat the result as the final probability. That quantity is only one arm’s unnormalized weight. It must be compared with arm 2’s weight and normalized:

P(A_{n+1}=1\mid\mathcal F_n) = \frac{\rho_n(\rho_n/q_{n,1})^\gamma} {\rho_n(\rho_n/q_{n,1})^\gamma+ (1-\rho_n)((1-\rho_n)/q_{n,2})^\gamma}

For example, in the first worked calculation the arm 1 weight is 0.900, not a probability of 0.900: after accounting for arm 2’s weight, the probability is $0.900/(0.900+0.2667)=0.7714$. Confusing a weight with a probability can produce values that do not sum to one across arms.

Key takeaway: The two-arm ratio-correction rule is obtained by weighting each arm’s target share by its target-to-current share ratio raised to $\gamma$, then normalizing across arms. Both arms’ weights are necessary to obtain the assignment probability.

Check Your Understanding

Try each question, then check the solutions below.

A hypothetical trial has $n=10$, with 2 assignments to arm 1 and 8 to arm 2. The arm 1 target is 0.50 and $\gamma=1$. Calculate both weights and the probability of assigning the next participant to arm 1.
In words, why is the arm 1 correction not determined by $\rho_n/q_{n,1}$ alone?
If the current shares equal their target shares and the target shares are positive, what probability does the formula assign to arm 1? Show the weight substitution.
Name two conditions under which the interior ratio-correction formula is not directly defined.

Solutions. (1) The current shares are $q_{n,1}=0.20$ and $q_{n,2}=0.80$. The weights are $w_{n,1}=0.50(0.50/0.20)=1.250$ and $w_{n,2}=0.50(0.50/0.80)=0.3125$. Thus $P(A_{11}=1\mid\mathcal F_{10})=1.250/(1.250+0.3125)=0.8000$. (2) The final probability is a normalized comparison of arm 1 and arm 2 weights; omitting arm 2’s weight does not account for the full set of assignment probabilities. (3) If $q_{n,1}=\rho_n$ and $q_{n,2}=1-\rho_n$, both ratios equal one. The weights are $w_{n,1}=\rho_n$ and $w_{n,2}=1-\rho_n$, so the arm 1 probability is $\rho_n/(\rho_n+1-\rho_n)=\rho_n$. (4) Examples include $n=0$, which makes current shares undefined, and a zero current share, which makes its target-to-current ratio undefined. A target share at zero or one also falls outside the stated interior conditions.

Approach	Arm 1 probability	Uses current allocation shares?
Direct target allocation	\(\rho_n\)	No
Ratio-correction DBCD	\(w_{n,1}/(w_{n,1}+w_{n,2})\)	Yes, through both arm-specific ratios